IPMAT Indore 2023 MCQ PYQs

Q1:ipmat indore 2023QA › Unit DigitEasyMCQ · MCQ

Let

p

be a positive integer such that the unit digit of

p ^ 3

is

4

. What are the possible unit digits of

(p + 3) ^ 3

A3
B1, 7, 9
C4, 7
D1, 3, 7

Pick an option to attempt

The Setup: We are given a positive integer

p

such that the unit digit of

p^3

is exactly

4

. We need to determine the possible unit digits of the expression

(p+3)^3

. Step 1: Determine the unit digit of

p

. We evaluate the cubes of all base digits from

0

to

9

to see which produces a unit digit of

4

:

0^3 = 0, 1^3 = 1, 2^3 = 8, 3^3 = 27, 4^3 = 64, 5^3 = 125, 6^3 = 216, 7^3 = 343, 8^3 = 512, 9^3 = 729

. The only digit whose cube ends in

4

is

4

itself. Therefore, the unit digit of

p

must be

4

. Step 2: Calculate the unit digit of

(p+3)^3

. Since

p \equiv 4 \pmod{10}

, we substitute this into the target expression:

p + 3 \equiv 4 + 3 \equiv 7 \pmod{10}

Now, cube this resulting unit digit:

7^3 = 343

The unit digit of

343

is

3

. Thus, the only possible unit digit for

(p+3)^3

is

3

. Final Answer: 3

Q2:ipmat indore 2023QA › Integral SolutionsMediumMCQ · MCQ

Let

[x]

denote the greatest integer not exceeding

x

and {

x

}

= x - [x]

If

n

is a natural number, then the sum of all values of

x

satisfying the equation

2[x] = x + n

{

x

} is

A $\dfrac{n(n+2)}{2}$
B $\dfrac{3}{2}$
C $\dfrac{n(n+1)}{2}$
D $n$

Pick an option to attempt

The Setup: We are given the equation

2[x] = x + n\{x\}

, where

n

is a natural number,

[x]

is the greatest integer function, and

\{x\}

is the fractional part. We must find the sum of all values of

x

that satisfy this equation. Step 1: Express

x

in terms of its integer and fractional parts. By definition, any real number

x

can be written as

x = [x] + \{x\}

. Substitute this into the given equation:

2[x] = ([x] + \{x\}) + n\{x\}

[x] = (n+1)\{x\}

Step 2: Isolate the fractional part and apply its bounds.

\{x\} = \frac{[x]}{n+1}

By the definition of the fractional part function,

0 \le \{x\} < 1

.

0 \le \frac{[x]}{n+1} < 1

Since

n

is a natural number (

n \ge 1

),

n+1

is strictly positive. Multiply the inequality by

n+1

:

0 \le [x] < n+1

Because

[x]

must be an integer, the possible values for

[x]

are the integers

0, 1, 2, \dots, n

. Step 3: Formulate

x

and calculate the sum. Substitute the expression for

\{x\}

back into

x = [x] + \{x\}

:

x = [x] + \frac{[x]}{n+1} = [x]\left(1 + \frac{1}{n+1}\right) = [x]\left(\frac{n+2}{n+1}\right)

To find the sum of all valid

x

, we sum this expression over all possible values of

[x]

from

0

to

n

:

\text{Sum} = \sum_{k=0}^{n} k\left(\frac{n+2}{n+1}\right) = \left(\frac{n+2}{n+1}\right) \sum_{k=0}^{n} k

Using the sum of the first

n

integers formula

\sum_{k=0}^n k = \frac{n(n+1)}{2}

:

\text{Sum} = \left(\frac{n+2}{n+1}\right) \left(\frac{n(n+1)}{2}\right) = \frac{n(n+2)}{2}

Final Answer:

\frac{n(n+2)}{2}

Q3:ipmat indore 2023QA › InequalitiesEasyMCQ · MCQ

The set of all real values of x satisfying the inequality

\dfrac{x^2(x+1)}{(x-1)(2x+1)^3} > 0

is

A $(-\infty, -1)$ U $(-\dfrac{1}{2}, 0)$ U $(1, +\infty)$
B $(-1, -\dfrac{1}{2})$ U $(1,+\infty)$
C $(-1, 0)$ U $(1,+\infty)$
D $(-1, -\dfrac{1}{2})$ U $(0,+\infty)$

Pick an option to attempt

The Setup: We must find the set of all real values of

x

that satisfy the rational inequality

\frac{x^2(x+1)}{(x-1)(2x+1)^3} > 0

. Step 1: Identify the critical points. Set the numerator and denominator factors to zero to find the critical points where the expression changes sign or becomes undefined: Numerator:

x = 0

(multiplicity 2, even),

x = -1

(multiplicity 1, odd). Denominator:

x = 1

(multiplicity 1, odd),

x = -1/2

(multiplicity 3, odd). The critical points in ascending order are:

-1, -1/2, 0, 1

. Step 2: Evaluate the sign intervals using a test-point method. Because the factor

x^2

is always non-negative and touches

0

at

x=0

, the sign does not change across

x=0

. We test the regions partitioned by the odd-multiplicity critical points (

-1, -1/2, 1

): * **Region

x > 1

:** Test

x = 2

. Expression is

(+)(+)/(+)(+) > 0

. (Valid) * **Region

-1/2 < x < 1

(excluding

0

):** Since sign alternates at

x=1

, this region is negative. (Invalid) * **Region

-1 < x < -1/2

:** Since sign alternates at

x=-1/2

, this region is positive. (Valid) * **Region

x < -1

:** Since sign alternates at

x=-1

, this region is negative. (Invalid) Step 3: Combine the valid intervals. The strict inequality requires the expression to be strictly greater than

0

, so the critical points themselves are not included. The valid intervals are

(-1, -1/2)

and

(1, +\infty)

.

\text{Set} = \left(-1, -\frac{1}{2}\right) \cup (1, +\infty)

Final Answer:

(-1,-\frac{1}{2})\cup(1,+\infty)

Q4:ipmat indore 2023QA › Profit & LossEasyMCQ · MCQ

A goldsmith bought a large solid golden ball at INR 1,000,000 and melted it to make a certain number of solid spherical beads such that the radius of each bead was one-fifth of the radius of the original ball. Assume that the cost of making golden beads is negligible. If the goldsmith sold all the beads at 20% discount on the listed price and made a total profit of 20%, then the listed price or each golden bead, in INR, was

A48000
B12000
C9600
D24000

Pick an option to attempt

The Setup: A solid golden ball costing INR

1,000,000

is melted into smaller spherical beads. The radius of each bead is one-fifth of the original ball's radius. The beads are sold at a

20\%

discount for a total overall profit of

20\%

. We must find the listed price per bead. Step 1: Calculate the number of beads produced. Let the radius of the original ball be

R

and the radius of a bead be

r

. We are given

r = \frac{R}{5}

. The volume of a sphere is proportional to the cube of its radius.

\text{Number of beads} = \frac{\text{Volume of original ball}}{\text{Volume of one bead}} = \frac{\frac{4}{3}\pi R^3}{\frac{4}{3}\pi r^3} = \left(\frac{R}{r}\right)^3 = (5)^3 = 125

So,

125

identical beads are produced. Step 2: Calculate the required revenue per bead. The total cost is INR

1,000,000

. The total profit made is

20\%

, so the total revenue generated is

120\%

of the cost:

\text{Total Revenue} = 1,000,000 \times 1.20 = 1,200,000

Since this revenue comes from selling

125

beads, the selling price (

SP

) per bead is:

SP = \frac{1,200,000}{125} = 9600

Step 3: Calculate the listed price. The selling price is derived after a

20\%

discount is applied to the listed price (

L

).

0.80 \times L = 9600

L = \frac{9600}{0.80} = 12000

Final Answer: 12000

Q5:ipmat indore 2023QA › CirclesMediumMCQ · MCQ

Which of the following straight lines are both tangent to the circle

x ^ 2 + y ^ 2 - 6x + 4y - 12 = 0

?

A $4x+3y+19=0, 4x+3y-31=0$
B $4x+3y-19=0, 4x+3y+31=0$
C $4x+3y-19=0, 4x+3y-31=0$
D $4x+3y+19=0, 4x+3y+31=0$

Pick an option to attempt

The Setup: We are asked to identify which pair of straight lines are both tangents to the circle defined by

x^2+y^2-6x+4y-12=0

. Step 1: Determine the center and radius of the circle. Complete the square for both

x

and

y

to convert the equation into standard circle form

(x-h)^2 + (y-k)^2 = r^2

:

(x^2 - 6x + 9) + (y^2 + 4y + 4) = 12 + 9 + 4

(x - 3)^2 + (y + 2)^2 = 25

The center of the circle is

C = (3, -2)

and the radius is

r = \sqrt{25} = 5

. Step 2: Apply the tangent distance formula. For a line

Ax + By + C = 0

to be tangent to a circle, the perpendicular distance from the circle's center to the line must exactly equal the radius

r

. Looking at the options, all candidate lines have the form

4x + 3y + c = 0

. Set the perpendicular distance from

(3, -2)

to the line equal to

5

:

\frac{|4(3) + 3(-2) + c|}{\sqrt{4^2 + 3^2}} = 5

\frac{|12 - 6 + c|}{\sqrt{25}} = 5

\frac{|6 + c|}{5} = 5

|6 + c| = 25

Step 3: Solve for the constant

c

. This absolute value equation yields two valid cases for

c

: Case 1:

6 + c = 25 \implies c = 19

Case 2:

6 + c = -25 \implies c = -31

The two tangent lines are

4x + 3y + 19 = 0

and

4x + 3y - 31 = 0

. Final Answer:

4x+3y+19=0,4x+3y-31=0

Q6:ipmat indore 2023QA › Matrices & DeterminantsMediumMCQ · MCQ

If

A = \begin{bmatrix} 1 & 2 \newline 3 & a \end{bmatrix}

where

a

is a real number and det

(A ^ 3 - 3A ^ 2 - 5A) = 0

then one of the values of

a

can be

A4
B6
C1
D5

Pick an option to attempt

The Setup: We are given a

2 \times 2

matrix

A

with a variable

a

. We must find a valid value for

a

given that

\det(A^3-3A^2-5A)=0

. Step 1: Factor the matrix polynomial.

A^3 - 3A^2 - 5A = A(A^2 - 3A - 5I)

Using the determinant product rule

\det(XY) = \det(X)\det(Y)

, the condition becomes:

\det(A) \cdot \det(A^2 - 3A - 5I) = 0

This means that either

\det(A) = 0

or

\det(A^2 - 3A - 5I) = 0

. Step 2: Test the first determinant condition. Calculate the determinant of matrix

A

:

A = \begin{bmatrix} 1 & 2 \\ 3 & a \end{bmatrix}

\det(A) = (1)(a) - (2)(3) = a - 6

If

\det(A) = 0

, then

a - 6 = 0 \implies a = 6

. Checking the multiple-choice options provided in the prompt,

6

is a listed option. Final Answer: 6

Q7:ipmat indore 2023QA › Time, Speed & DistanceEasyMCQ · MCQ

A helicopter flies along the sides of a square field of side length 100 kms. The first side is covered at a speed of 100 kmph, and for each subsequent side the speed is increased by 100 kmph till it covers all the sides. The average speed of the helicopter is

A250 kmph
B184 kmph
C192 kmph
D200 kmph

Pick an option to attempt

The Setup: A helicopter flies around the perimeter of a

100\text{ km} \times 100\text{ km}

square field. Its speed starts at

100\text{ kmph}

for the first side and increases by

100\text{ kmph}

for each subsequent side. We must calculate its average speed for the entire trip. Step 1: Determine the distance and speed for each leg. Since the field is a square with side length

100\text{ km}

, the helicopter flies four distinct legs of

100\text{ km}

each. * Side 1: Distance =

100\text{ km}

, Speed =

100\text{ kmph}

* Side 2: Distance =

100\text{ km}

, Speed =

200\text{ kmph}

* Side 3: Distance =

100\text{ km}

, Speed =

300\text{ kmph}

* Side 4: Distance =

100\text{ km}

, Speed =

400\text{ kmph}

Step 2: Calculate the time taken for each leg. Using Time = Distance / Speed: *

T_1 = \frac{100}{100} = 1\text{ hour}

*

T_2 = \frac{100}{200} = \frac{1}{2}\text{ hour}

*

T_3 = \frac{100}{300} = \frac{1}{3}\text{ hour}

*

T_4 = \frac{100}{400} = \frac{1}{4}\text{ hour}

Step 3: Calculate the average speed.

\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}}

\text{Total Distance} = 4 \times 100 = 400\text{ km}

\text{Total Time} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} = \frac{12 + 6 + 4 + 3}{12} = \frac{25}{12}\text{ hours}

\text{Average Speed} = \frac{400}{\frac{25}{12}} = 400 \times \frac{12}{25} = 16 \times 12 = 192\text{ kmph}

Final Answer: 192 kmph

Q8:ipmat indore 2023QA › Ratio, Proportion & VariationEasyMCQ · MCQ

\dfrac{(a + b)}{(b + c)} = \dfrac{(c + d)}{(d + a)}

which of the following statements is always true?

A $a + b + c + d = 0$
B $a=c,$ or $a+b+c+d = 0$
C $a=c$
D $a=c,$ and $b=d$

Pick an option to attempt

The Setup: We are given the algebraic equality

\frac{a+b}{b+c} = \frac{c+d}{d+a}

. We need to evaluate which conditional statement must necessarily follow. Step 1: Cross-multiply to clear the fractions.

(a+b)(a+d) = (c+d)(b+c)

Step 2: Expand both sides of the equation.

a^2 + ad + ab + bd = bc + c^2 + bd + cd

Step 3: Simplify and factor the expression. Subtract the right side from the left side. Notice that the

+bd

term exists on both sides and cancels out:

a^2 + ad + ab - bc - c^2 - cd = 0

Group the terms strategically to factor by grouping. Pair the squares, and group the remaining terms by common factors

b

and

d

:

(a^2 - c^2) + b(a - c) + d(a - c) = 0

Apply the difference of squares identity to the first term:

(a - c)(a + c) + b(a - c) + d(a - c) = 0

Factor out the common binomial term

(a - c)

:

(a - c)(a + c + b + d) = 0

Step 4: Evaluate the Zero Product Property. For this product to be zero, at least one of the factors must be zero: Factor 1:

a - c = 0 \implies a = c

Factor 2:

a + b + c + d = 0

Thus, it must always be true that

a=c

or

a+b+c+d=0

. Final Answer:

a=c,

or

a+b+c+d=0

Q9:ipmat indore 2023QA › PolynomialsMediumMCQ · MCQ

If the harmonic mean of the roots of the equation

(5 + \sqrt{2}) x ^ 2 - bx + 8 + 2\sqrt{5} = 0

is

4

then the value of

b

is

A $2$
B $4-\sqrt{5}$
C $3$
D $4+\sqrt{5}$

Pick an option to attempt

The Setup: We are given a quadratic equation

(5+\sqrt{2})x^2 - bx + 8+2\sqrt{5} = 0

whose roots have a harmonic mean of

4

. We must find the value of

b

. Step 1: Establish the Harmonic Mean formula for the roots. Let the roots of the quadratic equation be

\alpha

and

\beta

. The Harmonic Mean (HM) of two numbers is defined as:

HM = \frac{2\alpha\beta}{\alpha+\beta}

We are given that

HM = 4

. Step 2: Extract the sum and product of the roots from the quadratic. Using Vieta's formulas for

Ax^2 + Bx + C = 0

:

\text{Sum of roots } (\alpha+\beta) = -\frac{B}{A} = \frac{b}{5+\sqrt{2}}

\text{Product of roots } (\alpha\beta) = \frac{C}{A} = \frac{8+2\sqrt{5}}{5+\sqrt{2}}

Step 3: Substitute Vieta's results into the Harmonic Mean equation.

4 = \frac{2 \left( \frac{8+2\sqrt{5}}{5+\sqrt{2}} \right)}{\left( \frac{b}{5+\sqrt{2}} \right)}

Notice that the complex denominator

(5+\sqrt{2})

neatly cancels out from both the numerator and the denominator of the large fraction:

4 = \frac{2(8+2\sqrt{5})}{b}

4 = \frac{16+4\sqrt{5}}{b}

Step 4: Solve for

b

.

4b = 16 + 4\sqrt{5}

Divide both sides by

4

:

b = 4 + \sqrt{5}

Final Answer:

4+\sqrt{5}

Q10:ipmat indore 2023LRDI › TournamentsMediumMCQ · MCQ

In a chess tournament there are 5 contestants. Each player plays against all the others exactly once. No game results in a draw. The winner in a game gets one point and the loser gets zero points. Which of the following sequences cannot represent the scores of the five players?

A2, 2, 2, 2, 2
B3, 2, 2, 2, 1
C3, 3, 2, 1, 1
D4, 4, 1, 1, 0

Pick an option to attempt

The Setup: A round-robin chess tournament features

5

players where every player plays everyone else exactly once. A win awards

1

point, a loss awards

0

points, and no draws occur. We must identify which sequence of scores is mathematically impossible. Step 1: Establish the total points in the system. The number of players is

n = 5

. The total number of matches played in the tournament is

\binom{5}{2} = 10

. Since each match awards exactly

1

point to the winner and

0

to the loser, exactly

10

points are distributed across the 5 players. Every valid score sequence must sum to

10

. Step 2: Apply Landau's Theorem for score sequences. Landau's Theorem states that a sequence of

n

scores

s_1 \le s_2 \le \dots \le s_n

represents a valid round-robin tournament if and only if: 1. The sum of all scores equals

\binom{n}{2}

. 2. For any subset of

k

players (sorted from lowest to highest), their combined score must be at least

\binom{k}{2}

, representing the matches they played against each other.

\sum_{i=1}^{k} s_i \ge \binom{k}{2} \text{ for all } k=1, 2, \dots, n

Step 3: Evaluate the given options against Landau's criteria. Let's test the suspected invalid sequence

4, 4, 1, 1, 0

. Sort the sequence in ascending order:

s = (0, 1, 1, 4, 4)

. * Check

k=1

:

s_1 = 0 \ge \binom{1}{2} = 0

. (Passes) * Check

k=2

:

s_1 + s_2 = 0 + 1 = 1 \ge \binom{2}{2} = 1

. (Passes) * Check

k=3

:

s_1 + s_2 + s_3 = 0 + 1 + 1 = 2 \ge \binom{3}{2} = 3

. Here,

2 \ngtr 3

. The lowest three players must have generated at least

3

points purely from playing against each other, making a combined score of

2

physically impossible. Final Answer: 4, 4, 1, 1, 0

Q11:ipmat indore 2023QA › Simple & Compound InterestEasyMCQ · MCQ

If the difference between compound interest and simple interest for a certain amount of money invested for

3

years at an annual interest rate of

10\%

is INR

527

, then the amount invested in INR is

A17000
B15000
C1500
D170000

Pick an option to attempt

The Setup: The difference between Compound Interest (CI) and Simple Interest (SI) on a principal amount invested for

3

years at a

10\%

annual rate is INR

527

. We need to determine the original principal amount. Step 1: State the 3-year CI and SI difference formula. For a principal

P

invested for exactly

3

years at an annual interest rate

R

(expressed as a percentage), the difference

D

between CI and SI is given by the standard derived formula:

D = P \left(\frac{R}{100}\right)^2 \left(\frac{R}{100} + 3\right)

Step 2: Substitute the known values into the formula. We are given

D = 527

and

R = 10

.

527 = P \left(\frac{10}{100}\right)^2 \left(\frac{10}{100} + 3\right)

527 = P (0.1)^2 (0.1 + 3)

527 = P (0.01)(3.1)

527 = 0.031 P

Step 3: Solve for the Principal

P

.

P = \frac{527}{0.031} = \frac{527,000}{31}

Divide

527

by

31

to simplify the fraction:

527 \div 31 = 17

P = 17,000

Final Answer: 17000

Q12:ipmat indore 2023QA › Progression & SeriesMediumMCQ · MCQ

Let

a_{1}, a_{2}, a_{3}

be three distinct real numbers in geometric progression. If the equations

a_{1} x ^ 2 + 2a_{2}x + a_{3} = 0

and

b_{1} x ^ 2 + 2b_{2}x + b_{3} = 0

have a common root, then which of the following is necessarily true?

A $\dfrac{b_{1}}{a_{1}}, \dfrac{b_{2}}{a_{2}}, \dfrac{b_{3}}{a_{3}}$ are in geometric progression
B $b_{1}, b_{2}, b_{3}$ are in geometric progression
C $b_{1}, b_{2}, b_{3}$ are in arithmetic progression
D $\dfrac{b_{1}}{a_{1}}, \dfrac{b_{2}}{a_{2}}, \dfrac{b_{3}}{a_{3}}$ are in arithmetic progression

Pick an option to attempt

The Setup: Given three distinct real numbers

a_1, a_2, a_3

in a geometric progression (GP), we are told the quadratics

a_1x^2+2a_2x+a_3=0

and

b_1x^2+2b_2x+b_3=0

share a common root. We must deduce the relationship between the ratios of their coefficients. Step 1: Analyze the first quadratic equation. Since

a_1, a_2, a_3

are in GP, they satisfy the property

a_2^2 = a_1a_3

. Check the discriminant (

D

) of the first quadratic equation

a_1x^2 + 2a_2x + a_3 = 0

:

D = (2a_2)^2 - 4(a_1)(a_3) = 4a_2^2 - 4a_1a_3

Substitute

a_1a_3

for

a_2^2

:

D = 4(a_1a_3) - 4a_1a_3 = 0

Since

D = 0

, the first equation has a single repeated real root. We find this root using the quadratic formula:

x = \frac{-2a_2}{2a_1} = -\frac{a_2}{a_1}

Step 2: Apply the common root constraint. Because the two equations share a common root, and the first equation only has *one* unique root, this root (

x = -a_2/a_1

) must exactly be the root of the second equation

b_1x^2+2b_2x+b_3=0

. Substitute

x = -a_2/a_1

into the second equation:

b_1\left(-\frac{a_2}{a_1}\right)^2 + 2b_2\left(-\frac{a_2}{a_1}\right) + b_3 = 0

b_1\frac{a_2^2}{a_1^2} - 2b_2\frac{a_2}{a_1} + b_3 = 0

Step 3: Simplify the relationship to match the options. Divide the entire equation by

a_3

:

b_1\frac{a_2^2}{a_1^2 a_3} - 2b_2\frac{a_2}{a_1 a_3} + \frac{b_3}{a_3} = 0

Use the GP identity

a_2^2 = a_1a_3

to simplify the denominators. For the first term:

\frac{a_2^2}{a_1^2 a_3} = \frac{a_1 a_3}{a_1^2 a_3} = \frac{1}{a_1}

For the second term:

\frac{a_2}{a_1 a_3} = \frac{a_2}{a_2^2} = \frac{1}{a_2}

Substitute these back:

\frac{b_1}{a_1} - 2\frac{b_2}{a_2} + \frac{b_3}{a_3} = 0

2\left(\frac{b_2}{a_2}\right) = \frac{b_1}{a_1} + \frac{b_3}{a_3}

This fits the exact definition of an Arithmetic Progression (AP) where the middle term is the arithmetic mean of the outer terms. Final Answer:

\frac{b_{1}}{a_{1}}.

\frac{b_{2}}{a_{2}},

\frac{b_{3}}{a_{3}}

are in arithmetic progression

Q13:ipmat indore 2023QA › TrianglesHardMCQ · MCQ

In a triangle ABC, let D be the midpoint of BC, and AM be the altitude on BC. If the lengths of AB, BC and CA are in the ratio of 2:4:3, then the ratio of the lengths of BM and AD would be

A $11:4\sqrt{10}$
B $12:11$
C $11:12$
D $4\sqrt{10}:11$

Pick an option to attempt

The Setup: In

\triangle ABC

,

D

is the midpoint of

BC

, and

AM

is the altitude on

BC

. With side ratios

AB:BC:CA = 2:4:3

, we must find the length ratio

BM:AD

. Step 1: Assign algebraic lengths to the sides. Let the sides be

c = AB = 2x

,

a = BC = 4x

, and

b = CA = 3x

. Since

D

is the midpoint of

BC

,

BD = \frac{BC}{2} = 2x

. Step 2: Calculate the length of

BM

. Use the Law of Cosines to find

\cos B

in

\triangle ABC

:

\cos B = \frac{a^2 + c^2 - b^2}{2ac} = \frac{(4x)^2 + (2x)^2 - (3x)^2}{2(4x)(2x)}

\cos B = \frac{16x^2 + 4x^2 - 9x^2}{16x^2} = \frac{11x^2}{16x^2} = \frac{11}{16}

In the right-angled

\triangle ABM

, the segment

BM

represents the adjacent side to angle

B

. Thus,

BM = c \cdot \cos B

:

BM = 2x \left(\frac{11}{16}\right) = \frac{11x}{8}

Step 3: Calculate the length of the median

AD

. Apply Apollonius's Theorem to the median

AD

:

AB^2 + AC^2 = 2(AD^2 + BD^2)

(2x)^2 + (3x)^2 = 2(AD^2 + (2x)^2)

4x^2 + 9x^2 = 2AD^2 + 8x^2

13x^2 - 8x^2 = 2AD^2

5x^2 = 2AD^2 \implies AD^2 = \frac{5x^2}{2} \implies AD = \frac{x\sqrt{5}}{\sqrt{2}} = \frac{x\sqrt{10}}{2}

Step 4: Compute the ratio

BM : AD

.

\text{Ratio} = \frac{\frac{11x}{8}}{\frac{x\sqrt{10}}{2}} = \frac{11}{8} \times \frac{2}{\sqrt{10}} = \frac{22}{8\sqrt{10}} = \frac{11}{4\sqrt{10}}

Final Answer:

11:4\sqrt{10}

Q14:ipmat indore 2023QA › LogarithmsMediumMCQ · MCQ

Let

a, b, c

be real numbers greater than 1, and

n

be a positive real number not equal to 1. If

log_n(log_2a) = 1; log_n(log_2b) = 2

and

log_n(log_2c) = 3

then which of the following is true?

A $(b-a)^n=(c-b)$
B $a^n+b^n=c^n$
C $a+b=c$
D $(a^n+b)^n=ac$

Pick an option to attempt

The Setup: We are given a system of nested logarithms:

\log_n(\log_2 a)=1

,

\log_n(\log_2 b)=2

, and

\log_n(\log_2 c)=3

. We must establish which given algebraic relationship holds true. Step 1: Convert the logarithmic equations into exponential form. Using the rule

\log_x(y) = z \implies y = x^z

: Equation 1:

\log_2 a = n^1 = n \implies a = 2^n

Equation 2:

\log_2 b = n^2 \implies b = 2^{n^2}

Equation 3:

\log_2 c = n^3 \implies c = 2^{n^3}

Step 2: Express variables in terms of each other. Notice that the exponent of

b

is the square of the exponent of

a

:

b = 2^{n^2} = (2^n)^n = a^n

Notice that the exponent of

c

relates to

b

:

c = 2^{n^3} = (2^{n^2})^n = b^n

We also know that

c

relates to

a

by

c = (2^n)^{n^2} = a^{n^2}

. Additionally, multiply

a

and

c

:

ac = 2^n \cdot 2^{n^3} = 2^{n + n^3}

Alternatively,

ac = a \cdot b^n

. Step 3: Evaluate the multiple-choice options. We test the specific option

(a^n+b)^n = ac

to see if it holds true. Substitute

b = a^n

into the left side:

\text{Left Side} = (b + b)^n = (2b)^n

Distribute the exponent:

(2b)^n = 2^n \cdot b^n

Now, substitute our initial mappings back in. We know

2^n = a

and

b^n = c

:

2^n \cdot b^n = a \cdot c

The left side mathematically perfectly matches the right side

ac

. Final Answer:

(a^{n}+b)^{n}=ac

Q15:ipmat indore 2023QA › Permutation & CombinationEasyMCQ · MCQ

Consider an

8 \times 8

chessboard. The number of ways 8 rooks can be placed on the board such that no two rooks are in the same row and no two are in the same column is

A7
B7!
C8
D8!

Pick an option to attempt

The Setup: We need to find the total number of ways to place

8

indistinguishable rooks on a standard

8 \times 8

chessboard such that no two rooks threaten each other (no two share the same row or column). Step 1: Place the rooks sequentially row by row. To ensure no two rooks share a row, exactly one rook must be placed in each of the 8 rows. * Row 1: The first rook can be placed in any of the

8

squares (columns) in the first row. * Row 2: The second rook must be placed in the second row, but it cannot share the column occupied by the first rook. This leaves

7

valid squares. * Row 3: The third rook cannot share a column with the first two rooks, leaving

6

valid squares. *

\dots

* Row 8: The final rook is forced into the single remaining unoccupied column. Step 2: Calculate total configurations. The total number of valid placements is the product of the independent choices for each row:

\text{Total Ways} = 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 8!

Final Answer: 8!

Q16:ipmat indore 2023QA › Conic SectionsEasyMCQ · MCQ

The equation

x^2 + y^2 - 2x - 4y +5 = 0

represents

Aa pair of straight lines
Ba circle
Can ellipse
Da point

Pick an option to attempt

The Setup: We are given the conic equation

x^2+y^2-2x-4y+5=0

and asked to classify the specific geometric shape it represents. Step 1: Reformat the equation using completing the square. Group the

x

terms and

y

terms:

(x^2 - 2x) + (y^2 - 4y) = -5

Complete the square for

x

by adding

(-2/2)^2 = 1

: Complete the square for

y

by adding

(-4/2)^2 = 4

: Balance the equation by adding these to the right side as well:

(x^2 - 2x + 1) + (y^2 - 4y + 4) = -5 + 1 + 4

Step 2: Simplify and classify the equation.

(x - 1)^2 + (y - 2)^2 = 0

This is the standard form of a circle

(x-h)^2 + (y-k)^2 = r^2

. Here, the radius squared is exactly

0

(

r = 0

). A circle with a radius of

0

mathematically collapses into a single coordinate point located at its center,

(1, 2)

. Final Answer: a point

Q17:ipmat indore 2023QA › Progression & SeriesHardMCQ · MCQ

A person standing at the centre of an open ground first walks 32 meters towards the east, takes a right turn and walks 16 meters, takes another right turn and walks 8 meters, and so on. How far will the person be from the original starting point after an infinite number of such walks in this pattern?

A $32$ meters
B $\frac{32}{\sqrt{5}}$ meters
C $64$ meters
D $\frac{64}{\sqrt{5}}$ meters

Pick an option to attempt

The Setup: A person maps out a path taking consecutive

90^\circ

right turns, with each leg of the journey halving in distance (

32\text{m}, 16\text{m}, 8\text{m} \dots

). We must find their net displacement from the origin after infinite turns. Step 1: Establish a coordinate system. Let the starting point be the origin

(0,0)

. * Leg 1:

32\text{m}

East (Positive

x

-direction). * Leg 2:

16\text{m}

South (Negative

y

-direction). * Leg 3:

8\text{m}

West (Negative

x

-direction). * Leg 4:

4\text{m}

North (Positive

y

-direction). * Leg 5:

2\text{m}

East (Positive

x

-direction). This pattern continues infinitely. Step 2: Calculate the net displacement on the

x

-axis. The

x

-coordinate sequence alternates direction every two steps:

32

(East),

-8

(West),

2

(East),

-0.5

(West)

\dots

This forms an infinite geometric progression where the first term

a = 32

and the common ratio

r = -\frac{1}{4}

.

X_{\text{final}} = \frac{a}{1 - r} = \frac{32}{1 - (-1/4)} = \frac{32}{5/4} = 32 \times \frac{4}{5} = \frac{128}{5}

Step 3: Calculate the net displacement on the

y

-axis. The

y

-coordinate sequence begins on the second step:

-16

(South),

4

(North),

-1

(South)

\dots

This forms an infinite geometric progression where

a = -16

and

r = -\frac{1}{4}

. (We'll use magnitude for distance).

Y_{\text{final}} = \frac{16}{1 - (-1/4)} = \frac{16}{5/4} = 16 \times \frac{4}{5} = \frac{64}{5}

Step 4: Calculate the final straight-line distance. Use the Pythagorean theorem:

D = \sqrt{X_{\text{final}}^2 + Y_{\text{final}}^2}

.

D = \sqrt{\left(\frac{128}{5}\right)^2 + \left(\frac{64}{5}\right)^2} = \frac{64}{5} \sqrt{2^2 + 1^2} = \frac{64\sqrt{5}}{5}

Simplify by converting the denominator:

D = \frac{64\sqrt{5}}{(\sqrt{5}\cdot\sqrt{5})} = \frac{64}{\sqrt{5}}

Final Answer:

\frac{64}{\sqrt{5}}

meters

Q18:ipmat indore 2023QA › LogarithmsHardMCQ · MCQ

If

\log_{(cos x)}(sin x) + \log_{(sin x)}(cos x) = 2,

then the value of

x

is

A $n\pi + \dfrac{\pi}{4}, n$ is an integer
B $2n\pi + \dfrac{\pi}{4}, n$ is an integer
C $\dfrac{n\pi}{4}, n$ is an integer
D $\dfrac{n\pi}{4} + \dfrac{\pi}{4}, n$ is an integer

Pick an option to attempt

The Setup: We must solve for

x

given the trigonometric logarithmic equation

\log_{\cos x}(\sin x) + \log_{\sin x}(\cos x) = 2

. Step 1: Use substitution to solve the algebra. By the base-change inversion property of logarithms,

\log_a b = \frac{1}{\log_b a}

. Let

t = \log_{\cos x}(\sin x)

. The equation becomes:

t + \frac{1}{t} = 2

Multiply by

t

to form a quadratic:

t^2 - 2t + 1 = 0 \implies (t - 1)^2 = 0 \implies t = 1

Step 2: Re-substitute to find the trigonometric relation.

\log_{\cos x}(\sin x) = 1 \implies \cos^1 x = \sin x \implies \sin x = \cos x

Dividing by

\cos x

yields

\tan x = 1

. Step 3: Evaluate the domain restrictions for logarithms. For a logarithm

\log_b a

to be defined, the base

b > 0, b \neq 1

, and the argument

a > 0

. Thus, we strictly require

\sin x > 0

,

\sin x \neq 1

,

\cos x > 0

, and

\cos x \neq 1

. This strictly restricts valid solutions for

x

exclusively to the first quadrant of the unit circle. Step 4: Find the general solution for

x

. The first-quadrant angle where

\tan x = 1

is

\frac{\pi}{4}

. To represent all coterminal first-quadrant angles, we add full

360^\circ

rotations, which is

2n\pi

, where

n

is an integer.

x = 2n\pi + \frac{\pi}{4}

Final Answer:

2n\pi+\frac{\pi}{4},

n is an integer

Q19:ipmat indore 2023QA › Permutation & CombinationMediumMCQ · MCQ

A rabbit is sitting at the base of a staircase which has 10 steps. It proceeds to the top of the staircase by climbing either one step at a time or two steps at a time. The number of ways it can reach the top is

A144
B89
C34
D55

Pick an option to attempt

The Setup: A rabbit climbs a

10

-step staircase, taking either one or two steps at a time. We must find the total number of unique ways to reach the top. Step 1: Establish the recursive relationship. Let

W_n

be the number of ways to reach the

n

-th step. To reach step

n

, the rabbit must have either taken a single step from step

n-1

, or a double step from step

n-2

. Therefore, the total ways to reach step

n

is the sum of the ways to reach the previous two steps:

W_n = W_{n-1} + W_{n-2}

. This forms the Fibonacci sequence. Step 2: Define the base cases. * To reach Step 1 (

W_1

): Exactly

1

way (one 1-step). * To reach Step 2 (

W_2

): Exactly

2

ways (two 1-steps, or one 2-step). Step 3: Compute the sequence up to step 10. *

W_3 = W_2 + W_1 = 2 + 1 = 3

*

W_4 = 3 + 2 = 5

*

W_5 = 5 + 3 = 8

*

W_6 = 8 + 5 = 13

*

W_7 = 13 + 8 = 21

*

W_8 = 21 + 13 = 34

*

W_9 = 34 + 21 = 55

*

W_{10} = 55 + 34 = 89

Final Answer: 89

Q20:ipmat indore 2023QA › ProbabilityMediumMCQ · MCQ

The probability that a randomly chosen positive divisor of

10 ^ {2023}

is an integer multiple of

10 ^ {2001}

is

A $\dfrac{22}{2023}$
B $\dfrac{23^2}{{2024}^2}$
C $\dfrac{22}{2024}$
D $\dfrac{22^2}{{2023}^2}$

Pick an option to attempt

The Setup: We are asked for the probability that a randomly chosen positive divisor of

10^{2023}

is also an integer multiple of

10^{2001}

. Step 1: Determine the total number of divisors (the sample space). First, find the prime factorization of the base number:

10^{2023} = (2 \times 5)^{2023} = 2^{2023} \times 5^{2023}

The formula for the total number of divisors of

p^a \cdot q^b

is

(a+1)(b+1)

.

\text{Total Divisors} = (2023 + 1)(2023 + 1) = 2024^2

Step 2: Determine the number of valid target divisors. A divisor

D

is a multiple of

10^{2001}

(which is

2^{2001} \times 5^{2001}

) if its prime factorization

D = 2^x \times 5^y

meets the constraints: * For base 2:

2001 \le x \le 2023

* For base 5:

2001 \le y \le 2023

Calculate the number of integer choices for the exponents

x

and

y

: Choices for

x = 2023 - 2001 + 1 = 23

. Choices for

y = 2023 - 2001 + 1 = 23

.

\text{Number of valid multiples} = 23 \times 23 = 23^2

Step 3: Calculate the probability.

\text{Probability} = \frac{\text{Valid Multiples}}{\text{Total Divisors}} = \frac{23^2}{2024^2}

Final Answer:

\frac{23^{2}}{2024^{2}}

Q21:ipmat indore 2023QA › Set TheoryEasyMCQ · MCQ

In a group of 120 students, 80 students are from the Science stream and the rest are from the Commerce stream. It is known that 70 students support Mumbai Indians in the Indian Premier League; all the other students support Chennai Super Kings. The number of Science students who are supporters of Mumbai Indians is

AExactly 20
BBetween 15 and 25
C30 or more
DBetween 20 and 25

Pick an option to attempt

The Setup: A group of

120

students comprises

80

Science and

40

Commerce students. Exactly

70

students support Mumbai Indians (MI), while the rest support Chennai Super Kings (CSK). We must establish the possible range of Science students supporting MI. Step 1: Establish the basic parameters. Total Students =

120

Science Students =

80

Commerce Students =

120 - 80 = 40

Total MI Supporters =

70

Step 2: Calculate the maximum bound for Science MI supporters. Let

x

be the number of Science students who support MI. The maximum possible value for

x

occurs if as many MI supporters as possible are drawn from the Science pool. Since there are only

70

MI supporters total, and

80

Science students exist, it is mathematically possible for all

70

MI supporters to be Science students.

\text{Max}(x) = 70

Step 3: Calculate the minimum bound for Science MI supporters. The minimum possible value for

x

occurs if as many MI supporters as possible are drawn from the Commerce pool. There are only

40

Commerce students. If all

40

of them support MI, the remaining MI supporters must be Science students.

\text{Min}(x) = 70 (\text{Total MI}) - 40 (\text{Commerce MI}) = 30

Step 4: Conclude the range. The number of Science students supporting MI must sit exactly in the inclusive range

30 \le x \le 70

. Looking at the options, the constraint '30 or more' is the only logically correct bounding descriptor. Final Answer: 30 or more

Q22:ipmat indore 2023QA › Divisibility RulesEasyMCQ · MCQ

If a three-digit number is chosen at random, what is the probability that it is divisible neither by 3 nor by 4?

A $\frac{1}{4}$
B $\frac{1}{2}$
C $\frac{2}{3}$
D $\frac{1}{3}$

Pick an option to attempt

The Setup: We are asked for the probability that a randomly chosen

3

-digit number is divisible by *neither*

3

nor

4

. Step 1: Calculate the size of the sample space. The 3-digit numbers run inclusively from

100

to

999

.

\text{Total 3-digit numbers} = 999 - 100 + 1 = 900

Step 2: Use the Inclusion-Exclusion principle to find the complement. We first find the number of integers divisible by

3

OR

4

:

|A \cup B| = |A| + |B| - |A \cap B|

. * Divisible by 3: The sequence is

102, 105 \dots 999

.

\text{Count} = \frac{999 - 102}{3} + 1 = 297/3 + 1 = 300

* Divisible by 4: The sequence is

100, 104 \dots 996

.

\text{Count} = \frac{996 - 100}{4} + 1 = 896/4 + 1 = 225

* Divisible by 12 (Intersection): The sequence is

108, 120 \dots 996

.

\text{Count} = \frac{996 - 108}{12} + 1 = 888/12 + 1 = 75

\text{Divisible by 3 or 4} = 300 + 225 - 75 = 450

Step 3: Calculate the target probability. The target numbers are those *not* divisible by

3

or

4

.

\text{Target count} = \text{Total} - (\text{Divisible by 3 or 4}) = 900 - 450 = 450

\text{Probability} = \frac{450}{900} = \frac{1}{2}

Final Answer:

\frac{1}{2}

Q23:ipmat indore 2023QA › ProbabilityEasyMCQ · MCQ

The minimum number of times a fair coin must be tossed so that the probability of getting at least one head exceeds 0.8 is

A5
B6
C3
D7

Pick an option to attempt

The Setup: We must find the minimum number of independent coin tosses required for the probability of getting at least one head to exceed

0.8

. Step 1: Formulate the probability equation. The probability of getting *at least one* head is the complement of getting *zero* heads (all tails).

P(\text{at least 1 H}) = 1 - P(\text{all T})

For a fair coin tossed

n

times, the probability of getting all tails is

\left(\frac{1}{2}\right)^n

.

P(\text{at least 1 H}) = 1 - \left(\frac{1}{2}\right)^n

Step 2: Set up and solve the inequality. We require this probability to strictly exceed

0.8

.

1 - \left(\frac{1}{2}\right)^n > 0.8

1 - 0.8 > \left(\frac{1}{2}\right)^n

0.2 > \left(\frac{1}{2}\right)^n

Convert the decimal to a fraction (

0.2 = \frac{1}{5}

):

\frac{1}{5} > \frac{1}{2^n}

Inverting the fractions flips the inequality:

2^n > 5

Step 3: Test integer values for

n

. * If

n = 1

:

2^1 = 2 \ngtr 5

* If

n = 2

:

2^2 = 4 \ngtr 5

* If

n = 3

:

2^3 = 8 > 5

(Valid) The minimum integer number of tosses required is

3

. Final Answer: 3

Q24:ipmat indore 2023QA › TrigonometryMediumMCQ · MCQ

If

cos \alpha

+

cos \beta

= 1 then the maximum value of

sin \alpha - sin \beta

is

A $\sqrt{2}$
B $2$
C $\sqrt{3}$
D $1$

Pick an option to attempt

The Setup: Given the constraint

\cos\alpha + \cos\beta = 1

, we must find the maximum value of

\sin\alpha - \sin\beta

. Step 1: Rewrite both expressions with the sum-to-product identities. Let

u = \frac{\alpha+\beta}{2}

and

v = \frac{\alpha-\beta}{2}

. The constraint and the target expression become:

\cos\alpha + \cos\beta = 2\cos u \cos v = 1

\sin\alpha - \sin\beta = 2\cos u \sin v

Both share the same factor

2\cos u

, which is what makes the constraint substitutable. Step 2: Eliminate

u

using the constraint. The constraint gives

2\cos u = \frac{1}{\cos v}

directly (note

\cos v \neq 0

). Substitute this into the target:

\sin\alpha - \sin\beta = \left(\frac{1}{\cos v}\right)\sin v = \tan v

The whole problem has collapsed to maximising

\tan v

. Step 3: Find the admissible range of

v

.

\cos u

is a cosine, so it is bounded by

|\cos u| \le 1

:

\left|\frac{1}{2\cos v}\right| \le 1 \implies |\cos v| \ge \frac{1}{2}

On the branch where

\tan v

is positive and increasing, this restricts

v

to

0 \le v \le \frac{\pi}{3}

. Step 4: Maximise on that range.

\tan v

increases throughout

\left[0, \frac{\pi}{3}\right]

, so the maximum sits at the right endpoint:

\text{Maximum} = \tan\frac{\pi}{3} = \sqrt{3}

Step 5: Confirm the maximum is actually attained. At

v = \frac{\pi}{3}

the constraint forces

\cos u = \frac{1}{2\cos(\pi/3)} = 1

, so

u = 0

, giving

\alpha = \frac{\pi}{3}

and

\beta = -\frac{\pi}{3}

. Check both conditions:

\cos\frac{\pi}{3} + \cos\left(-\frac{\pi}{3}\right) = \frac{1}{2} + \frac{1}{2} = 1

\sin\frac{\pi}{3} - \sin\left(-\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2} = \sqrt{3}

Both hold, so

\sqrt{3}

is genuinely achieved and is the maximum. Final Answer:

\sqrt{3}

Q25:ipmat indore 2023QA › RemainderMediumMCQ · MCQ

A polynomial

P(x)

leaves a remainder

2

when divided by

(x - 1)

and a remainder

1

when divided by

(x - 2)

The remainder when

P(x)

is divided by

(x - 1)(x - 2)

is

A $3-x$
B $3$
C $x-3$
D $2$

Pick an option to attempt

The Setup: A polynomial

P(x)

leaves specific remainders when divided by two linear binomials. We must use the Polynomial Remainder Theorem to deduce the linear remainder when divided by their quadratic product. Step 1: Establish the given values using the Remainder Theorem. The Remainder Theorem states that dividing

P(x)

by

(x-a)

yields a remainder of

P(a)

. * Divided by

(x-1)

, remainder is

2 \implies P(1) = 2

. * Divided by

(x-2)

, remainder is

1 \implies P(2) = 1

. Step 2: Formulate the Division Algorithm equation. When

P(x)

is divided by a quadratic polynomial

(x-1)(x-2)

, the maximum possible degree of the remainder is linear. Let the remainder be

R(x) = ax + b

.

P(x) = Q(x)(x-1)(x-2) + (ax + b)

Step 3: Substitute the known

x

values to create a system of equations. Substitute

x = 1

:

P(1) = Q(1)(0)(-1) + (a(1) + b) \implies a + b = 2

Substitute

x = 2

:

P(2) = Q(2)(1)(0) + (a(2) + b) \implies 2a + b = 1

Step 4: Solve the linear system for

a

and

b

. Subtract the first equation from the second equation:

(2a + b) - (a + b) = 1 - 2

a = -1

Substitute

a

back into the first equation:

-1 + b = 2 \implies b = 3

The resulting linear remainder is

R(x) = -1x + 3 = 3 - x

. Final Answer:

3-x

Q26:ipmat indore 2023LRDI › ArrangementsEasyMCQ · MCQ

A pharmaceutical company has tested five drugs on three different organisms. The following incomplete table reports if a drug works on the given organism. For example, drug A works on organism R while B and C work on Q.

Drug	Organism Q	Organism R
A		Y
B	Y
C	Y
D
E

Following additional information is available: Each drug works on at least one organism but not more than two organisms. Each organism can be treated with at least two and at most three of these five drugs. On whichever organism A works, B also works. Similarly, on whichever organism C works. D also works. D and E do not work on the same organism.

show less

Drug E works on

AP & R
BQ & R
COnly P
DOnly R

Pick an option to attempt

The Setup: A pharmaceutical company tested five drugs (A, B, C, D, E) on three organisms (P, Q, R). We must deduce the complete relationship matrix using the provided table and logical constraints. Step 1: Establish the initial matrix and apply direct implication rules. From the provided table, A works on R, B works on Q, and C works on Q. Rule 3 states that if A works on an organism, B also works on it (

A \implies B

). Since A works on R, B must also work on R. Rule 4 states that if C works on an organism, D also works on it (

C \implies D

). Since C works on Q, D must also work on Q. Step 2: Apply column capacity constraints. Rule 2 states each organism can be treated with at most 3 drugs. Currently, Organism Q is treated by B, C, and D, reaching its absolute maximum capacity. Therefore, neither A nor E can work on Q. Since A is restricted from Q and must act as a strict subset of B's placements (which are only Q and R), A must solely work on R. Step 3: Apply disjoint rules to isolate P and R. Rule 2 dictates Organism P needs at least 2 drugs. The only available candidates for P are C, D, and E (A and B are locked into Q and R). Rule 5 states D and E cannot work on the same organism. Thus, P cannot be treated by both D and E. To secure 2 drugs for P without pairing D and E, P must logically be treated by C and D (choosing C automatically forces D via Rule 4). Step 4: Finalize Drug E's placement. Drug E cannot work on Q (capacity full) and cannot work on P (disjoint with D). Rule 1 mandates each drug works on at least 1 organism. Thus, E is forced to work on Organism R. The finalized matrix is: P={C,D}, Q={B,C,D}, R={A,B,E}. Drug E strictly works only on organism R. Final Answer: Only R

Q27:ipmat indore 2023LRDI › ArrangementsEasyMCQ · MCQ

A pharmaceutical company has tested five drugs on three different organisms. The following incomplete table reports if a drug works on the given organism. For example, drug A works on organism R while B and C work on Q.

Drug	Organism Q	Organism R
A		Y
B	Y
C	Y
D
E

Following additional information is available: Each drug works on at least one organism but not more than two organisms. Each organism can be treated with at least two and at most three of these five drugs. On whichever organism A works, B also works. Similarly, on whichever organism C works. D also works. D and E do not work on the same organism.

show less

Drug D works on:

AP & Q
BOnly P
COnly Q
DQ & R

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The Setup: Using the fully deduced matrix from the pharmaceutical constraints, we need to determine the exact organisms treated by Drug D. Step 1: Reference the derived matrix. As mathematically established by the constraint deduction: P={C,D}, Q={B,C,D}, R={A,B,E}. Step 2: Isolate Drug D. Evaluating the assignments, Drug D successfully operates on both Organism P and Organism Q. Final Answer: P & Q

Q28:ipmat indore 2023LRDI › ArrangementsEasyMCQ · MCQ

A pharmaceutical company has tested five drugs on three different organisms. The following incomplete table reports if a drug works on the given organism. For example, drug A works on organism R while B and C work on Q.

Drug	Organism Q	Organism R
A		Y
B	Y
C	Y
D
E

Following additional information is available: Each drug works on at least one organism but not more than two organisms. Each organism can be treated with at least two and at most three of these five drugs. On whichever organism A works, B also works. Similarly, on whichever organism C works. D also works. D and E do not work on the same organism.

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The organism(s) that can be treated with three of these five drugs is(are)

AOnly P
BOnly Q
CP and Q
DQ and R

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The Setup: We need to identify which organisms are treated by exactly three of the five drugs, relying on the completely deduced matrix. Step 1: Count the drugs assigned to each organism. According to the derived matrix: Organism P is treated by C and D (Total: 2). Organism Q is treated by B, C, and D (Total: 3). Organism R is treated by A, B, and E (Total: 3). Step 2: Select the qualifying organisms. Both Q and R meet the criteria of being treated by exactly three drugs. Final Answer: Q and R

Q29:ipmat indore 2023LRDI › ArrangementsEasyMCQ · MCQ

A pharmaceutical company has tested five drugs on three different organisms. The following incomplete table reports if a drug works on the given organism. For example, drug A works on organism R while B and C work on Q.

Drug	Organism Q	Organism R
A		Y
B	Y
C	Y
D
E

Following additional information is available: Each drug works on at least one organism but not more than two organisms. Each organism can be treated with at least two and at most three of these five drugs. On whichever organism A works, B also works. Similarly, on whichever organism C works. D also works. D and E do not work on the same organism.

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Organism R can be treated with

AA, B and C
BOnly A and E
COnly A and B
DA, B and E

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The Setup: We must list the specific combination of drugs that successfully treat Organism R based on the logical constraints. Step 1: Reference Organism R in the derived matrix. As established in the base matrix deduction: P={C,D}, Q={B,C,D}, R={A,B,E}. Step 2: Isolate Organism R. The column for Organism R strictly contains treatments from Drugs A, B, and E. Final Answer: A, B and E

Q30:ipmat indore 2023LRDI › ArrangementsEasyMCQ · MCQ

A pharmaceutical company has tested five drugs on three different organisms. The following incomplete table reports if a drug works on the given organism. For example, drug A works on organism R while B and C work on Q.

Drug	Organism Q	Organism R
A		Y
B	Y
C	Y
D
E

Following additional information is available: Each drug works on at least one organism but not more than two organisms. Each organism can be treated with at least two and at most three of these five drugs. On whichever organism A works, B also works. Similarly, on whichever organism C works. D also works. D and E do not work on the same organism.

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Organism P can be treated with:

AOnly C and D
BB, C and D
COnly B and D
DA, B, and D

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The Setup: We must list the specific combination of drugs that successfully treat Organism P based on the logical constraints. Step 1: Reference Organism P in the derived matrix. As established in the base matrix deduction: P={C,D}, Q={B,C,D}, R={A,B,E}. Step 2: Isolate Organism P. To satisfy the minimum 2-drug requirement while avoiding the D/E conflict constraint, Organism P is treated exclusively by Drugs C and D. Final Answer: Only C and D

IPMAT Indore 2023 — MCQ

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